Question

Find the area of the region included between y = x² + 3 and the line y = x + 3.

Answer 1

The given parabola is y = x² + 3, i.e. (x – 0)² = y – 3

∴ Its vertex is P(0, 3)

To find the points of intersection of the line and the parabola

Equating the values of y from both the equations, we get

x² + 3 = x + 3

∴ x² – x = 0

∴ x(x – 1) = 0

∴ x = 0 or x = 1

When x = 0, y = 0 + 3 = 3

When x = 1, y = 1 + 3 = 4

∴ The points of intersection are P(0, 3) and B(1, 4)

Required area = area of the region PABCP

= (area of the region OPABDO) – (area of the region OPCBDO)

Now, area of the region OPABDO

= area under the line y = x + 3 between x = 0 and x = 1

= `int_0^1 y  dx, "where"  y = x + 3`

= `int_0^1 (x + 3) dx`

= `int_0^1x  dx + 3 int_0^1 1  dx`

= `[x^2/2]_0^1 + 3[x]_0^1`

= `(1/2 - 0) + 3(1 - 0)`

= `1/2 + 3`

= `7/2`

Area of the region OPCBDO

= area under the parabola y = x² + 3 between x = 0 and x = 1

= `int_0^1 y  dx, "where"  y = x^2 + 3`

= `int_0^1 (x^2 + 3)dx`

= `int_0^1 x^2  dx + 3 int_0^1 1  dx`

= `[x^3/3]_0^1 + 3[x]_0^1`

= `(1/3 - 0) + 3(1 - 0)`

= `1/3 + 3`

= `10/3`

∴ Required area = `7/2 - 10/3`

= `(21 - 20)/(6)`

= `1/6 "sq units"`.